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\begin{document}
\title{Introduction to Algebraic K-Theory}
\author{Mona Merling}
\date{}
\maketitle
Due to its historical development, algebraic $K$-theory has two main components: the study of lower $K$-groups ($K_0$, $K_1$, and $K_2$), which have explicit algebraic descriptions used in applications, and the study of higher $K$-theory, which was introduced by Quillen.
Quillen gave two different constructions for the higher $K$-groups - the $+$ construction and the $Q$-construction - and proved the``$+=Q$'' theorem, which states that his two definitions agree. The first definition was particularly useful in computations, while the second one was more appropriate for proving theorems.
\section*{$K_0$}
Let $M$ be an abelian monoid. Then we can define the Grothendieck group $Gr(M)$ to be the free abelian group on $M$ modulo the relation $[m]+[n]=[m+n]$. For example, $Gr(\N) = \Z$.
Recall that topological $K$-theory (defined by Atiyah and Hirzebruch in 1959) sets the complex $K$-theory group to be $$KU^0(X) = Gr(\Vect_\C(X)),$$ where $\Vect_\C(X)$ is the monoid of (isomorphism classes of) complex vector bundles over $X$, with addition given by Whitney sum of bundles. Similarly, real $K$-theory was defined as $$KO^0(X) = Gr(\Vect_\R(X)).$$
\begin{definition}
The $0^{th}$ $K$-group of a ring is defined as $$K_0(R) = Gr(\mc{P}(R)),$$ where $\mc{P}(R)$ is the monoid of isomorphism classes of finitely-generated projective $R$-modules under direct sum.
\end{definition}
\begin{exercise}
Why do we require finite generation? (Hint: Eilenberg swindle.)
\end{exercise}
\begin{example}
Let $F$ be a field. Then $\mc{P}(F) \cong \N$ since every module is free, so $K_0(F)=\Z$.
\end{example}
\begin{example}
Let $R$ be a PID. By the structure theorem for modules over a PID, we know that projective modules are free. Thus again $K_0(R)=\Z$.
\end{example}
\begin{example}
Let $K$ be a number field, and let $\mc{O}_K$ be its number ring (a/k/a ring of integers). It's a theorem that every number ring is a Dedekind domain, and the structure theorem for finitely-generated modules over Dedekind domains tells us that if $P$ is projective of rank $n$, then $P \cong \mc{O}_K^{n-1} \oplus I$, where $I$ is a uniquely defined class in the class group $Cl(K)$. Thus $K_0(\mc{O}_K) = \Z \oplus Cl(K)$.
\end{example}
\begin{example}
If $X$ is a compact Hausdorff space, then $KU^0(X) \cong K_0 (C^0(X,\C))$. (Here $C^0$ denotes continuous functions.) This is called \textit{Swan's theorem}.
\end{example}
There are many other applications. For example, \textit{Wall's finiteness obstruction} is an algebraic-K-theoretic obstruction to a CW complex being homotopy equivalent to a finite CW complex.
\section*{$K_1$}
If $R$ is a ring, we can define $GL(R) = \colim_n GL_n(R)$ along the inclusions $GL_n(R) \hookrightarrow GL_{n+1}(R)$ given by inclusion into the upper-left corner (with a 1 in the bottom-right corner and zeros elsewhere). We then define $$K_1(R) = GL(R)/[GL(R),GL(R)].$$
\begin{lemma}[Whitehead]
$[GL(R),GL(R)] = E(R)$, the elementary matrices.
\end{lemma}
If $R$ is commutative, then $\det:GL(R) \ra R^\times$ induces a map $\det:K_1(R)\ra R^\times$. The kernel of this map is denoted $SK_1(R)$. On the other hand, we can include $R^\times \hookrightarrow K_1(R)$ via the identification $R^\times \cong GL_1(R)$. This splits the determinant map, and hence $K_1(R) \cong R^\times \oplus SK_1(R)$.
\begin{example}
Let $F$ be a field. In this case, $E(F)=SL(F)$, and hence $K_1(F)=F^\times$.
\end{example}
\begin{example}
Let $K$ be a number field and $\mc{O}_K$ be its number ring. By a theorem of Bass-Milnor-Serre, $SK_1(\mc{O}_K)=0$. Thus $K_1(\mc{O}_K)\cong \mc{O}_K^\times \cong \mu(K)\oplus \Z^{r_1+r_2-1}$. The latter isomorphism is by the Dirichlet unit theorem, $\mu(K)$ denotes the unit group of $K$, $r_1$ is the number of real embeddings of $K$, and $r_2$ is the number of conjugate pairs of complex embeddings of $K$.
\end{example}
\section*{Higher $K$-groups}
First we give some motivation for studying higher $K$-theory.
\subsection*{Quillen-Lichtenbaum Conjecture}
Recall the (analytic version of the) class number formula:
\begin{equation*}
\lim_{s\ra 0} \frac{\zeta_K(s)}{s^{r_1+r_2-1}} = - \frac{h_K}{w_K}R_K,
\end{equation*}
where
\begin{itemize}
\item $\zeta_K(s)$ is the Dedekind $\zeta$-function (if $K=\Q$, this is just the Riemann $\zeta$-function);
\item $h_K$ is the class number of $K$;
\item $w_K$ is the number of roots of unity in $K$;
\item $R_K$ is the regulator of $K$.
\end{itemize}
Observe that $h_K = |K_0(\mc{O}_K)_{tor}|$, while $w_K = |K_1(\mc{O}_K)_{tor}|$.
\begin{conjecture}[Quillen-Lichtenbaum]
\begin{equation*}
\lim_{s\ra -m} \frac{\zeta_K(s)}{(s+m)^{rk(K_{2m+1}(\mc{O}_K))}} = \pm \frac{ |K_{2m}(\mc{O}_K)|}{K_{2m+1}(\mc{O}_K)_{tor}}
\end{equation*}
\end{conjecture}
This gives an indication that higher $K$-groups can have extremely deep implications.
\subsection*{Vandiver's Conjecture}
We give another example of the deep connections between number theory and algebraic $K$-theory. The proof of the following conjecture would be a landmark in algebraic number theory because it would imply other important conjectured results.
\begin{conjecture}[Vandiver's conjecture]
Let $K=Q(\zeta_p)^+$ be the maximal real subfield of the $p$-th cyclotomic field $\Q(\zeta_p)$. Then $p$ does not divide the class number of $K$.
\end{conjecture}
It has been shown that Vandiver's conjecture is equivalent to the statement that $K_{4i}(\Z)=0$ for all $i$. This should seem quite amazing.
\subsection*{Definition of higher K-groups}
Recall that complex topological $K$-theory is a generalized cohomology theory given by the 2-periodic spectrum $KU = \{BU, \Omega BU, BU, \Omega BU, \ldots \}$, while real topological $K$-theory is given by the 8-periodic spectrum $KO=\{BO,\ldots, \Omega^8 BO, \ldots \}$. Note that as topological groups, $O \simeq GL(\R)$ and $U\simeq GL(\C)$. By analogy with topological $K$-theory, one might try to define higher algebraic $K$-groups for a ring $R$ using the space $BGL(R)$, where again, $BGL(R)$ stands for the classifying space of the group $GL(R)$. However, now $GL(R)$ is a discrete group, so $BGL(R)$ is a $K(GL(R),1)$. This suggests one needs to make a modification to the construction, and Quillen defined the higher $K$-groups by using the Quillen $+$-construction of $BGL(R)$: $$K_i(R)=\pi_i(BGL(R)^+)\text{ for } i\geq 1,$$ where the +-construction is taken with respect to the commutator subgroup of $GL(R)$. Recall that the plus-construction kills a perfect subgroup of the fundamental group and is an isomorphism on homology.
\begin{definition}
We define the \textit{K-theory space} to be $K(R) = K_0(R) \times BGL(R)^+$, and then for all $i \geq 0$ we can set $K_i(R) = \pi_i(K(R))$.
\end{definition}
Let's check: $$K_1(R) = \pi_1(BGL(R)^+) = \pi_1(BGL(R))/[GL(R),GL(R)] = GL(R)/[GL(R),GL(R)],$$ so the new definition is consistent witht he previous definition of $K_1$.
\subsection*{Relation of the +-construction to group completions}
There are many ways to get the +-construction, and they all yield homotopy equivalent spaces. One way (e.g.\! the way Hatcher does it) is by attaching 2- and 3-cells. A more conceptual way to obtain the space $BGL(R)^+$ is as the basepoint component of the ``group completion'' of the $H$-space $\coprod_{n=0}^\infty BGL_n(R)$, and this is a key point in the proof of the $+=Q$ theorem.
We recall the following definition of a group completion of a space. Recall that for a homotpy-associative and -commutative $H$-space $X$ $\pi_0(X)$ is an abelian monoid, $H_0(X;R)$ is the monoid ring $R[\pi_0(X)]$, and the integral homology $H_*(X;R)$ is an associative graded-commutative ring with unit.
\begin{definition}
Let $X$ be a homotopy-commutative and -associative H-space. A map $f:X\ra Y$, where $Y$ is a homotopy-commutative and -associative H-space, is called a \textit{group completion} if:
\begin{itemize}
\item $\pi_0(Y) = Gr(\pi_0(X))$, and
\item $H_*(Y;R)= \pi_0(X)^{-1} H_*(X,R)$ for any coefficients $R$.
\end{itemize}
\end{definition}
The idea in equivariant algebraic $K$-theory is to replace the classifying bundles for $GL_n(R)$ in $\coprod BGL_n(R)$ with equivariant bundles.
Quillen's motivation behind this definition of higher $K$-groups was to compute the $K$-groups of finite fields, which he was able to do by describing $BGL(R)^+$ as the homotopy fiber of a computable map.
\section*{The Q-construction}
There was a need to prove ``fundamental theorems of $K$-theory'' that had already been defined for $K_0$ and $K_1$, such as the additivity, the resolution, the devissage, and the localization theorems. However, Quillen wasn't able to do this using the definition we've already seen. So instead he gave a new definition using the Q-construction, generalized the theorems, and then proved the $+=Q$ theorem, i.e., that the two definitions agree.
We will use the notion of \textit{exact categories} (e.g.\! abelian categories, additive categories with split-exact sequences (e.g.\! $\mc{P}(R)$)). We will denote monomorphisms by $\hookrightarrow$ and epimorphisms by $\twoheadrightarrow$.
\begin{definition}
Let $\mc{C}$ be an exact category. We construct the category $Q\mc{C}$ with $\ob(Q\mc{C})=\ob(\mc{C})$ and with $Q\mc{C}(A,B)$ consisting of isomorphism classes of diagrams $A \twoheadleftarrow C \hookrightarrow B$.
\end{definition}
\begin{definition}
The \textit{classifying space} of a category $\mc{C}$ is given by the composition
\begin{equation*}
\cat{Cat} \stackrel{\textup{Nerve}}{\ra} \cat{sSet} \stackrel{|-|}{\ra} \cat{Top}.
\end{equation*}
Functors of categories turn into maps of classifying spaces, and natural transformations turn into homotopies of maps.
\end{definition}
\begin{exercise}
Why do natural transformations correspond to homotopies?
\end{exercise}
\begin{exercise}
Compute the classifying space of the category with one object and two morphisms (where the non-identity morphism squares to the identity). (Hint: This is the group $\Z/2$, thought of as a category.)
\end{exercise}
\begin{definition}
Given an exact category $\mc{C}$, we define $K(\mc{C})=\Omega B(Q\mc{C})$, and then we set $$K_i(\mc{C}) = \pi_i(K(\mc{C})) = \pi_{i+1}(BQ\mc{C}).$$
\end{definition}
\section*{$+=Q$}
\begin{theorem}[+=Q]
For a ring $R$, $$\Omega BQ\mc{P}(R) \simeq K_0(R) \times BGL(R)^+.$$
\end{theorem}
The proof is incredibly involved, and would probably take two 45-minute talks to expose. The strategy of the proof is be to define an intermediate object, the category $S^{-1}S$ associated to a certain symmetric monoidal category $S$, and to show that $B(S^{-1}S)$ is homotopy equivalent to each of the terms in the equivalence we are trying to prove.
\end{document}